# 4232 solve using equations 4210 and 427 the radial

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42.32. Solve: Using Equations 42.10 and 42.7, the radial probability density for the 1 s state is ( ) ( ) B 2 2 2 2 2 1 1 4 4 r a r s s P r r R r A r e π π = = This radial probability density peaks at the point where dP r /dr = 0. The derivative is B B B 2 2 2 2 2 2 1 1 B B 2 4 2 8 1 r a r a r a r s s dP r r A re e A re dr a a π π = = The derivative is zero at r = a B , so P r ( r ) is a maximum at this value of r .
42.33. Solve: (a) From Equation 42.7, the 2 p radial wave function is ( ) B 2 2 2 B 2 p r a p A R r re a = The graph of R 2 p ( r ) is seen to have a single maximum. (b) R 2 p ( r ) is a maximum at the point where dR 2 p / dr = 0. The derivative is B B B 2 2 2 2 2 2 B B B B 1 2 2 2 2 p p p r a r a r a dR A A r r e e e dr a a a a = = The derivative is zero at r = 2 a B , so R 2 p ( r ) is a maximum at r = 2 a B . (c) The radial wave function R 2 p and the probability density P 2 p ( r ) is smaller at r = 4 a B than it is at r = 2 a B . We are less likely to find the electron at a point with r = 4 a B than at a point with r = 2 a B . However, there are many more points with r = 4 a B than with r = 2 a B . The increased number of points more than compensates for the decreased probability per point. As a result, it is more probable to find the electron at distance r = 4 a B than it is at distance r = 2 a B .

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42.34. Solve: (a) For a hydrogen atom in the p -state, l = 1. Because L z = m = and 1 2 , z S = ± = the three possible values of L z are = , 0, and = , and the two possible values of S z are 1 2 = and 1 2 . = We can now compute J z using J z = L z + S z . The results are presented in the following table. L z S z J z m j = 1 2 + = 3 2 = 3 2 = 1 2 = 1 2 = 1 2 0 1 2 + = 1 2 = 1 2 0 1 2 = 1 2 = 1 2 = 1 2 + = 1 2 = 1 2 = 1 2 = 3 2 = 3 2 (b) The values of J z found above can be divided into two groups: ( ) 3 3 1 1 2 2 2 2 , , , = = = = and ( ) 1 1 2 2 , = = . Because J z = m j = and m j = j to j , the j values of the above two groups are 3 2 and 1 2 .
42.35. Solve: The electron configuration in the ground state of K ( Z = 19) is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 1 This means that all the states except the 4 s state are completely filled. For Ti ( Z = 22), all the states up to the 4 s state are completely filled. The electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 2 The d subshell has only two electrons. In the case of Fe ( Z = 26), the 3 d subshell has six electrons and all the lower level states are completely filled. The electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 The ground-state electron configurations of Ge ( Z = 32) and Br ( Z = 35) are 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 5

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42.36.
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